Electrical Power, Energy, and Joule Heating

What you’ll learn

Calculate electrical energy transfer and resistive heating.

គណនាអានុភាព ថាមពលអគ្គិសនី និងកម្ដៅជូល។

Meaning and use

Electrical power is voltage times current for the steady direct-current model used here. For a constant resistor, Ohm’s law gives equivalent I squared R and V squared over R forms. Energy equals power times time when power is constant or an appropriate average. Resistive electrical energy can become heat; other devices can have additional outputs.

ក្នុងគំរូចរន្តជាប់ថេរ P=VI។ សម្រាប់រេស៊ីស្តង់ថេរ អាចប្រើ I²R ឬ V²/R។ ថាមពលស្មើអានុភាពគុណពេល ហើយថាមពលក្នុងរេស៊ីស្តង់អាចផ្ទេរជាកម្ដៅ។

Formulas

Steady DC power

\[P=VI=I^2R=\frac{V^2}{R}\]

Energy and ideal resistive heating

\[E=Pt=VIt,\qquad Q=I^2Rt\]

Symbols and units

Symbol	Meaning	Unit or restriction
P	Electrical power	W
V, I, R	Voltage, current, and positive resistance	V, A, Ω
t	Elapsed time	s for energy in J
E, Q	Electrical energy and heat transfer in the stated resistor model	J

Conditions and limits

• The I²R and V²/R forms use the resistor relation and appropriate constant values.
• These examples use steady DC; do not apply the simple product blindly to arbitrary AC waveforms.
• កុំសន្មតថាថាមពលអគ្គិសនីទាំងអស់ជាកម្ដៅមានប្រយោជន៍ក្នុងឧបករណ៍គ្រប់ប្រភេទ។

Worked examples

Power and energy

A component has 12 V across it and carries 2 A for 30 s.

1. Power \(P=12\times2=24\) W.
2. Energy \(E=24\times30=720\) J.

24 W; 720 J

អានុភាព និងថាមពលមានឯកតាខុសគ្នា។

A resistor’s heating

A constant 5 Ω resistor carries 2 A for 10 s.

1. Power \(I^2R=2^2\times5=20\) W.
2. Heating energy \(Q=20\times10=200\) J.

200 J

ចរន្តត្រូវលើកការ៉េ មុនគុណ R និង t។

Common mistakes

Instead of	Use	Why
P = V/I.	P = VI.	V/I gives resistance, not power.
Q = IRt for resistor heating.	Q = I²Rt.	The heating expression includes squared current.

Practice

Solve these before revealing the worked solutions. Include units where relevant.

សាកល្បងដោះស្រាយមុនបើកមើលចម្លើយ។ សូមបង្ហាញជំហាន និងឯកតាដែលពាក់ព័ន្ធ។

1. Find power for V = 6 V and I = 0.5 A.
2. Find energy for constant 10 W over 60 s.
3. Find resistor heating for I = 3 A, R = 2 Ω, t = 5 s.

Answers and steps

Show solutions / បង្ហាញដំណោះស្រាយ

1. 3 W

   6 × 0.5 = 3.

2. 600 J

   10 × 60 = 600.

3. 90 J

   3² × 2 × 5 = 90.

Quick recap

• Use a stated electrical model and consistent values.
• Distinguish power from energy.
• Square current in I²Rt.